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I have a sequence of functions given by

  1. $f_n(x) = x^n$, on $D=(0,1)$

  2. $f_n(x) = \begin{cases} 4n^2x , & 0 \leq x \leq \frac{1}{2n} \\ -4n^2x+4n ,& \frac{1}{2n} \leq x \leq \frac{1}{n} \\ 0 & \text{otherwise} \end{cases}$, on $D=[0,1]$.

How can I find a sequence $(x_n)$ in $D$ such that $$\lim_{n\rightarrow\infty} f_n(x_n) \neq 0 = \lim_{n\rightarrow\infty} f(x_n) ?$$

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  • $\begingroup$ Is $f$ the pointwise limit of $(f_n)$? $\endgroup$ – BigbearZzz Mar 23 '16 at 9:54
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Look at where uniform convergence "breaks"; for $f_n$, this happens at $1$, and if you let $x_n=1-\frac{1}{n}$, you find that $$f_n(x_n)=\left(1-\frac{1}{n}\right)^n\to\frac{1}{e}\neq 0.$$ For the second function it is helpful to draw the graph; for fixed $n$ we get a piecewise linear function, and it has its maximum at $\frac{1}{2n}$. Setting $x_n=\frac{1}{2n}$, we have that $$f_n(x_n)=1\to 1\neq 0.$$

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